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Design Oxygen Analyzer
Please help me with this molecular Formula question?
Carbohydrates, an important source of good energy, are compounds that contain carbon, hydrogen, and oxygen. A food chemist extracts a carbohydrate from honey and submits a sample to a spectroscopy lab for anylasis.Complete the Anaylsis section of the lab report.
Experimental Design
A sample of the carbohydrate is burned in a combustion analyzer to determine the percent by mass of each element in the compund. Another sample is anaylized by a mass spectrometer to determine the molar mass of the carbohydrate.
Evidence
percent by mass of carbon = 40.0%
percent by mass of hydrogen = 6.8%
percent by mass of oxygen = 53.2%
molar mass= 180.2g/mol
Analysis
a) Calculate the molecular formula of the carbohdrate.
1. Assume 100 grams of substance. So these percentages should be represented as 40.0 g, 6.8 g and 53.2 g.
Convert them to moles.
40.0 g /12.0 g/mol = 3.33 mol
6.8 g/ 1.01 g/mol = 6.73 mol
53.2 g/16.0 g/mol= 3.33 mol
2. Divide by the smallest number of moles.
3.33 mol/3.33= 1
6.73 mol/3.33= 2
3.33 mol/3.33= 1
3. Write out the empirical formula using the results obtained as subscripts.
Therefore, CH2O.
4. To determine the molecular formula,
a. First get the mass of the empirical formula. It’s 30.0 g.
b. Divide the molar mass by the empirical formula mass. The answer is 6.
c. Multiply this answer to the coefficients in the empirical formula.
6(CH2O) = C6H12O6
The molecular formula of the carbohydrate is C6H12O6, or glucose.
Continuous, Fixed-position Biogas / Methane Analysis, Monitoring & Measurement